text:
Would the combustion chambers of a turbine engine generate any thrust by itself?
Well, I know that the combustion chambers drive the compressors, and the reaction occurs, and all that stuff.
But let's imagine that somehow someone is converting a turbine engine (turbojet, turboprop, or whatever) into a Hero's engine "aeolipile" where the combustion chambers are used to rotate the entire device—and thus, the compressor—with its chemical reaction.

Of course, all of this would be a thesis on its own, so let's just concentrate on the question in the title regarding combustion chambers.

If someone where to expose the combustion chambers of a turbine to the open air without the turbine, still suplying compressed air and fuel in the right measurements, would the expanding gases exiting combustion chambers produce any thrust?

It depends on what you mean by "by itself".  The diagram provided in the link by @fooot below does a nice job of breaking down the forces per component in an engine -- but that is still an engine with everything working together.  I.e. there is still a compressor providing lots of flow to the combustor.  If a combustor really was 'by itself', it would be little more than a candle flame.

You have basically invented the ramjet.

@tylisirn absolutely, just make the exhaust area slightly larger (to accommodate the heated gas) and it will generate thrust.  But it needs an external power source to get it going. This one ran on economy grade gasoline.

Yes you can.  However you would need a means to compress the air prior to it entering the combustion chamber.  This can be accomplished with a turbo compressor as in a typical jet engine. Other methods are to use stagnation pressure at the air inlet, which is what a ramjet engine uses.
A third option, which has not been pursued since the early days of jet aircraft is the motorjet. This is a jet engine, which uses an auxiliary engine to drive the compressor.  Very few aircraft or designers ever pursued this, but one exception was the Italian made Caproni Campini N.1 aircraft developed in the 1930s.

This design was effective, but with the advent of better metallurgical technology and the development of the turbo supercharger (which is, essentially a gas turbine engine sans the isobaric combustion chamber), the motorjet fell out of favor compared to turbojets due to an inferior thrust to weight ratio.

Yes. The expanding gases from combustion is what does the work in an engine. Most of the thrust of a turbojet engine comes from the combustors. The rest of the engine is really just there to provide the stream of compressed air for the combustion process.

Technically yes, BUT...
...The compressor of a jet engine is one of very few devices capable of actually supplying the required amount of air at the required pressure.
According to PEter's answer here - How much power is drawn by a turbofan to power its own compressor? - a 200 kN engine can be expected to draw 95 MW at the compressor at takeoff. Most of this power is then recovered, and the engine produces a peak output of 40-50 MW at the low-pressure shaft.
There are a few wind tunnels in the world that can produce this kind of power. A smaller engine, or the same engine at cruise, would require less. Still, you're looking at a major industrial installation to supply the air. Or at another jet engine.
Jet fuel does not ignite easily, it will not do so at ambient pressure. Commonplace air compressors will not provide enough pressure or enough flow. If you take a really small engine, like one used on radio-controlled jet models, which could be supplied, then the combustion chamber will produce some amount of thrust.

supplying fuel and air in the right measurements, would the expanding gasses exiting the combustion chambers produce any thrust?

Exposed to the open air, no.  Essentially you've built yourself a fireplace.  The circulation would be at best be cooler denser air being drawn in from the bottom.  Hotter, less dense air would rise.  Adding fuel in the "right proportions" you have ... a Bunsen burner.
The key is raising pressure in the combustion chamber to increase both mass and velocity of exiting gasses in order to generate meaningful thrust:
$ mass × velocity/second$

Thrust is a Force = mv/s = ma = kg × meters/second$^2$

Back to Heros Engine.  This is a steam engine.  Pressure is raised by using an external heat source to expand liquid water into steam.  The expansion ratio is 1 to 1700.
Heros engine spins by expelling steam vapor out its nozzles at high velocity.  But significant pressure is required to do this. Action - Reaction. (Tea kettles won't fly, but steam boilers can explode).
Back to our jet engine, along with its rocket and reciprocating (piston) relatives.  These are internal combustion engines, which provide thrust from a heat source inside a "combustion chamber", creating pressure.
One little problem is that generating extremely high pressures and expelling high velocity gasses can be terribly inefficient compared to converting pressure into mechanical energy to turn a wheel, or propeller.

the quantity of fuel required to produce a given amount of thrust for a given amount of time is call Thrust Specific Fuel Consumption

It's inverse is ... Isp or Specific Impulse.
Rockets are the worst, but will function in any environment.  So will Hero's engine.  If you are ever stuck 10 million miles from earth with a nuclear reactor and happen to find an icey asteroid, you're going home!
Reciprocating engines create cylinder pressure in power pulses to generate mechanical energy, jets create power continuously to generate mechanical energy by spinning a turbine.

Jets have Horsepower!

Jet need their "Horsepower" to increase incoming air pressure in the combustion chamber by turning a compressor.  This is called a "turbojet".  In spite of being less efficient, turbojets are able to generate thrust at speeds that mechanical devices such as propellers cannot match.

unfortunately, these days, fuel is expensive

So now jet engines use their "Horsepower" to turn not only compressors, but also their fans, which are able to generate thrust at higher speeds than propellers.


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How do you know the maximum load a control surface can withstand?

source
This is a separate strength testing (load testing) of the elevator of an airplane, how was the load determined? How did the engineers determine the maximum elevator load?

Which of the 3 questions are we answering here? Basically,  I'm going with "test to destruction", and when it fails the maximum load has been determined.

That wasn't the question tho of the 3 questions.  That answer requires more info than is available to answer. That would have to come from the designer.  How much is needed? Enough to provide full elevator authority when coming in engine out and be able to safely flare.

Elevator loads are the function of the design and flight envelope. Looks like a small plane, so look up for example CS-VA 391 (or the correct national code for your category) how to calculate the loads. Base air force distribution on the horizontal tail can usually be assumed to be triangular, and increment from control surface deflection need to be added, also usually approximated as a triangular distribution acting from the hinge line.

The elevator must be able to withstand aero loads at $V_{NE}$ and maximum deflection, using a safety factor of 1.5 at least.
The aero load = the hinge moment $H_M$ is:
$$H_M = C_h \cdot ½ \rho {V_h}^2 \cdot S_s \cdot {\bar{c}}_s$$
with

$½ \rho {V_h}^2$ = dynamic pressure at the control surface
$S_s$ = control surface area
${\bar{c}}_s$ = mean aerodynamic chord of the surface behind the hinge axis.
$C_h$ = factor depending on angle of attack and surface deflection. The figure below is depicts the coefficients for the Fokker 27. (From TU Delft D 26-1 by prof. Gerlach)

Can you Please provide the reference for the graph and the equation you shared?

Re "The elevator must be able to withstand aero loads at VNE and maximum deflection, using a safety factor of 1.5 at least." -- really?  Why bother to design the elevator to withstand a condition that will inevitably fail the wing?

The graph is from university lecture handouts, I don't have access to them right now and cannot look up the exact figure. The equation is in the lectures as well.

@quietflyer Really. Wing structural failures are one thing, preventing the elevator from failing while the pilot is attempting to recover from a full stall is another highly recommended structural feature.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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text:
When do pilots perform a CAT II/III autoland?
If I understand correctly in good, visual weather get conditions airline pilots generally do ILS CAT I, when at some point they disconnect autopilot at some point and land manually. If the visibility is low, they do ILS of higher category (II/III)  with autoland.
My question is, what are the criteria that pilots decide they will be shooting autoland? Is it RVR below CAT I minima or below some higher limit, just to be sure? Also, does ceiling matter in making this decision? For example OVC001, or OVC002 with good RVR and visibility - is it then required to autoland?

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How to transform a NACA airfoil into a near-circle using the inverse Karman-Trefftz?
Is there anyway to find the leading-edge's center of curvature of a NACA Airfoil?
I'm trying to transform a NACA airfoil into a near-circle using the inverse Karman-Trefftz transformation. the transformation has 2 singularity points, s1 which is located on the trailing edge, and s0 is located midway between the leading edge nose and its center of curvature.. now for symmetrical airfoils it's easy to find s0, since the center of curvature is on the x axis, and the leading edge radius formula is known, but for cambered airfoils it's a bit complicated with MATLAB, I've done all the hard work, and now stuck here.

Possible duplicate of What does leading edge radius mean?

Leading-edge radius r=1.1019(t), where t is the airfoil thickness.... but how can i find the center of the circle in the leading edge if i only know the radius ??

As you say yourself, you know the radius of curvature on the leading edge. You also probably have the leading-edge coordinate itself. You appear to have the equations for the airfoil, too ($x(t), z(t)$). That means you can also compute the tangent at the leading edge:
$X_{LE}=\begin{pmatrix}
x_{LE}\\
z_{LE}
\end{pmatrix}$
$\dot X_{LE}=\begin{pmatrix}
dx/dt\\
dz/dt
\end{pmatrix}_{LE}$
You might also get the tangent directly from the profile parameters, in the same way you got the curvature radius. Note I'm regarding the airfoil in (x,z) space, where x points downstream and z upwards, as is custom in aircraft aerodynamics and flight mechanics (y is the spanwise direction).
The center of the leading-edge circle is on a normal line to the leading-edge tangent, at the leading edge, at a distance of $r_{LE}$:
$X_{cLE} = X_{LE} +
\begin{pmatrix} dz/dt\\-dx/dt \end{pmatrix}_{LE}
{{r_{LE}}\over{||\dot X_{LE}||}}$
In the general case, you may not have $r_{LE}$ directly, but you can of course compute it from the parametric equations(or after-the-fact numerical  analysis of a given shape, using finite differences). Instead of typing the equations for curvature radius out, I'll just refer you to Wikipedia for this:
https://en.wikipedia.org/wiki/Curvature#In_terms_of_a_general_parametrization


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Is it possible to know why an airline operates specific equipment on certain routes?
I've spent far too long staring at flight tracking sites, and occasionally I think I'll notice some pattern in what types of aircraft a given airline uses for certain scenarios/routes.
For example, Delta seems to operate a lot of MD-88's on short-haul routes that other airlines would likely use ERJs or CRJs on. They also seem to use 757s on quite short hops (ATL to southeastern cities ~1hr away).
Another example would be why British Airways operates 747s to some cities, and 777s to others, and exactly how they decide where to use which aircraft.
Is there a good reference for figuring out the reasoning behind why airlines use a given aircraft for a given route? Or are there too many different factors and too much corporate secrecy to really know for sure?

Without sending them an email and asking I presume?

It all comes down to economics. If  Passenger income is >  Plane cost (long term leases) + fuel cost + crew cost + meals costs + terminal fees + ground crew fees + overflight fees + overhaul/inspections + indirect costs  ... then the flight is making money.   If not, then one way to make a profit is to change the plane & payload.  Or, eat the loss and make it up on other routes to keep the lossy route open, say to feed the larger planes flying from a hub for long distances from smaller out of the way airports.  I'm sure a large corporation like BA has this all modelled and tracked very well.

@ratchetfreak I mean, that is a very logical option I hadn't considered. Definitely worth a shot!

An airline I once worked for regularly flew half empz767s on a route otherwise operated by a narrowbody because the extra underbelly cargo paid for it all. So no, often it’s not obvious...

@CptReynolds I dream of finding a ticket on one of those flights where the plane is far larger than it need to be (with respect to just passengers). I've noticed 777s flying DTW-ATL, and I can't imagine those are very full.

@zymhan Those 777s (and sometimes A350s, too) flying DTW-ATL are actually probably pretty close to full most of the time. That's a hub-hub route for Delta between the largest airline hub in the world (ATL) and Delta's primary hub for trans-Pacific flights. Even when it's not operated by a widebody, it's operated by Delta's largest narrow-body aircraft, such as 757-300s, 757-200s, and A321s. That being said, the primary reason for a widebody on that route is almost certainly just to get the equipment to the other city to operate a long-haul flight.

As far as Delta operating 757s (and even 767s) regularly on short hops within the Southeast, this is mainly to cities where DL needs a lot of seats and frequency is already high. ATL-MCO traffic has grown to the point where during certain times of day, they fly a 757, A321, or 737-900ER every half hour (and it's hourly for most of the rest of the day.) ATL-BNA operates 180-200 seats roughly hourly on narrowbodies (and they're usually full in my experience being a Delta frequent flyer based at BNA.)

There are many factors, but you can often predict what class of plane will be used for a particular route by the distance and demand.
For smaller cities, they want to fill the planes, but they also need several flights per day to cover the fixed costs of serving that airport (and make things convenient for customers), so it's often better to have a commuter jet bouncing to the nearest hub and back several times per day day than one large plane that carries an entire day's worth of passengers in one shot.  For larger non-hub cities, they'll do roughly the same thing but with mid-size planes.
Between hubs, they'll run the largest narrow-bodies to connect all the various small cities not served by a common hubs, plus hubs themselves tend to be large cities with plenty of native demand too.
For international, they'll run wide-bodies because they need higher fuel capacity, and there's rarely a convenient time for customers due to the long flight times combined with crossing several time zones, so you might as well pack them into as few planes as possible.
Notably, some carriers fly several different types, each optimized for specific route types, while others fly just a few types (or even just one) because that makes maintenance, pilot training and fleet scheduling easier--but may limit what kinds of routes they can fly profitably.  When airlines merge, they'll usually have similar fleets--and in cases where they don't, they'll quickly start dumping types they don't like and replacing them with types they do.

Every airline has their own recipe and priorities in this regard and it’s unlikely you will find any open accessible references to this.
Apart from the obvious demand, availability and costs of aircraft and business they do take in consideration the wear and tear (and maintenance cycles) of the aircraft (how long to go before what maintenance) as well as a balance of long and short haul operations and the specifics of each particular aircraft and its parts. An aircraft with some inoperative equipment might still be legal and dispatched for a short haul flight but not ETOPs operations.
It can go down to small details like customer profile and preferences as well as configuration of the aircrafts interior (eg. newly upgraded seats) and the availability of service and maintenance facilities.

That makes sense. Is aircraft maintenance related to hours, or cycles, or both? I imagine if some parts are cycle-bound, and some are hour-bound, that you might end up with a mismatch over time, where the craft has flown relatively few longer-distance flights, or vice versa.

Depending on the items it could be: calendar time, landings, operating cycles, flight hours or block times.
Some parts are checked on condition maintenance which is a periodic inspection or testing of  the components or structures against a known standard or wear limit.

A major German carrier used to fly 747s on German domestic routes (for reference, if required - it’s hard to squeeze more than 600NM into a German domestic route) at one point before their retirement, allegedly because the airframes were coming up to their next scheduled base maintenance input on hours, but not yet on cycles.

You also get the occasional really big planes on routes that wouldn't seem to necessarily make sense for the distance.  This can happen when there's a big out and back that brings the plane back to the hub with some "downtime".  For example, out of SFO, a large plane may fly a 8 hour leg out to a far point, and then come back.  Adding time in for servicing etc, the plane would arrive back at the hub site around 18 hours after it left.  It could serve the same route and just sit on the ground for 6 hours waiting, but why not send it back and forth to DEN for example, which it could do in that downtime.  Or maybe the timing works to have the plane go to the east coast and then do a long trip out of there.

This makes sense! I've noticed much larger than normal flights operating between two international hubs, presumably because it was needed for an international flight from the destination city.

You can find this kind of stuff out on flightaware using the "track inbound plane" function.  I just started playing with UA340 which is a 777 IAD -> SFO and just keep hitting 'track inbound' to see where this tail number has been.  Back far enough we get HKG -> ORD : ORD -> EWR : EWR -> FRA : FRA -> ORD :ORD -> FRA : FRA -> IAD :IAD -> SFO and presumably SFO to a long haul destination.

Maintenance is also a factor here: the airline may route aircraft to a particular hub where they have maintenance facilities. If a UA 777 ends up in IAD and is due for maintenance at the base in SFO, they can either use it for a revenue flight or ferry it empty, so why not try to fill it up?

Another reason for this: ETOPS is not allowed for the first flight after maintenance, so a large twin may need to fly one domestic flight before it's legal to return to international service.

@StephenS I'd never heard that rule before!


text:
How can skydivers start at 15,000 feet without the jumper suffering from hypoxia?
Many commercial skydiving facilities offer jumps at 15,000 ft. However, when I took my ground theory exams, it was reiterated many times that hypoxia starts at 10,000 feet.
How does this work? Is it because of the short time spent at that altitude?

At 15k, unless you're a smoker, you won't really get hypoxic symptoms at 15k ft.  Your night vision will be degraded mostly. 10000 ft is used to provide lots of fudge factor to account for variations in people, and those with poor lung capacity.  La Paz is almost 13000 ft and you can acclimatize to that.  Pilots in WW1 would patrol at 20000 ft w/o oxygen.  Crazy mountain climbers go higher than that, with the really crazy ones scaling Everest w/o oxy, after years of special training. When you are at high alt just sitting, you can't tell the difference unless you are exerting yourself.

Climbers living at sea level routinely climb Mt Rainier at 14,410' without oxygen or any significant high elevation acclimatization other than normal conditioning hikes in the lower mountains averaging 3000' to 7000'ish... I've had cabin pressure bleed off slowly to 17-18k and started feeling symptoms, but they were mild enough that I recognized them and caught it.

People routinely climb 14,000'+ mountains without oxygen. Even when excluding Alaska, the U.S. has more than 60 such peaks.

You wouldn't use oxygen to climb Mont Blanc at 15,780ft, but I know from personal experience that people will experience the oxygen deprivation in a number of ways. In my case it was just physical (making very slow progress) but others in the party were affected mentally, e.g. making poor decisions and losing their temper.

@JohnK, at 15k, you'll make mistakes at about five times the rate you would at sea level.

They don't. Hypoxia is an actual concern in medium and high altitude skydiving.
It has been specifically researched as a possible safety risk. Skydivers jumping from over 10,000 ft do begin to exhibit mild symptoms of hypoxia - reduced ability to concentrate, less situational awareness, etc - even after a short time at that altitude. It's usually not a problem on its own, but can increase the risk of human error.
To reduce this concern, skydivers use supplemental oxygen onboard the aircraft to keep their blood oxygen-saturated as long as possible. But once it's disconnected, the symptoms start to build up.
Hypoxia starts subtle, somewhat similar to light inebriation, and takes a while (or a very low ppO2) to reach full blackout. Still, in rare cases, loss of consciousness does happen in skydiving.
A full blackout incident has been recorded in this video. You can see the use of the supplementary oxygen system until right before the jump, but two of the skydivers had to share one oxygen mask. One of them passes out from hypoxia in freefall, and is rescued by his buddies. This is a 22K jump - the higher the altitude and the longer the exposure, the worse the symptoms.

However, there are permanent human settlements above 15,000 ft. When living at low pressure for a while, the human body will acclimatize to lower oxygen pressure. Virtually anyone can do it, but individual tolerance varies. A healthy lifestyle (mostly no smoking), good fitness (lung capacity, aerobic exercise), and specific training regimens can greatly accelerate this acclimatization.
The risk of blackouts from hypoxia is mitigated in just a couple days of training close to the target altitude. Long-term acclimatization goes on to also eliminate the mental symptoms, so after weeks or months a person can function continuously at training altitude at nearly full capacity.
Experienced skydivers acclimatize and develop tolerance to hypoxia. Just one jump a day over a few weeks does the same work as the couple days of conditioning hikes for mountain climbers. Doing any of your first (or first few dozen) jumps from 15,000 ft might not be a good idea, unless you've lived in the mountains. It's not so much the blackout risk as the secondary risks from the mental effects of hypoxia.
Acclimatization requires time; it doesn't happen during one flight. Multiple jumps in one day, or prolonged loitering before the jump will make the symptoms worse, as shown in linked research. Jumps spread over consecutive days will let the body adapt and reduce the symptoms.

Something is up. I do not easily get hypoxia above 14,000 feet. I will eventually have problems that high up, but it develops over hours of exertion. I'm having a hard time reconciling the 15,000 feet extreme hypoxia with my 14,400 I can exert myself and feel nothing wrong for the first hour.

@Joshua If you've done multiple sessions of multiple hours of exerting exercise at 14,000 ft, you've probably already acclimatized. That's the LLTH training regimen from the linked article.

When I went tandem skydiving the professional skydivers would have little tubes of oxygen they could stick in their mouths until about 30 seconds before they jumped. Since their passengers didn't need to be so alert, we were allowed to be mildly hypoxic (probably helped calm some of the more nervous ones!)

You're right that we're taught that hypoxia might become an issue above 10,000ft but there is no magic line. It's not like at 9,999ft you're fine then BAM 10,000ft you're in trouble. Hypoxia affects different people differently, often affected by general health, smoking and other physiological factors.
That said, you're assumption is almost certainly right, it's the lack of time spent in that zone that makes it relatively safe.
A quick calculation shows that at terminal velocity, a human will fall from 15,000ft to 10,000ft in roughly 30 seconds. According to SKYbrary:

Hypoxia occurs within a few minutes if the cabin pressure altitude rises to between 5,000-6,000 m (about 16,000 - 20,000 ft).

So with 30seconds in the "Hypoxia zone" (as it were) it's not really a problem. Any higher than 15,000ft there is good guidance that supplementary oxygen is an option:

For intermediate-altitude jumps (15,000-20,000 feet MSL), participants should hold at least a USPA B license and have made 100 jumps.
For high-altitude jumps (20,000-40,000 feet MSL), participants should:

hold a USPA C license
have made at least one jump from 15,000 feet MSL or below using the same functioning bailout oxygen system

For extreme-altitude jumps (40,000 feet MSL and higher), participants should:

hold a USPA D license
have made at least two jumps from below 35,000 feet MSL using the same functioning bailout oxygen and pressure systems

Unless you skydive on the way up, you're going to be spending a bit more than 30s in the low oxygen environment ;)

@Sanchises It's a good point, but even the climb from under 10K to 15K isn't particularly long.

One could be breathing o2 on the way up, too, @Sanchises.

10k to 15k , lineup, preparations is an easy 5 minute time.

I think the danger here is being overblown quite a bit.
About a million people a year visit the top of Pikes Peak, at 14,112 feet, so not quite 15,000, but well over 10,000. They do keep oxygen bottles at the top of the peak, but mostly to sell to tourists at hugely inflated prices, not because a significant number of people really need them as medical treatment.
It is fairly common for visitors to suffer from altitude sickness--feeling somewhat lightheaded and sometimes a bit nauseous. Quite a few have a little trouble sleeping the first few days as well. But that's rarely any sort of acute episode where their life is in any danger. Typically it's more like their plans change from getting up at 4 AM and being on the hiking trail by 5, into something like going to the hotel restaurant for brunch around 10 or 11. They mostly just lose ambition, not their lives.
FAA requirements aren't based so much on the idea that at 12 or 15 thousand feet you will suffer from hypoxia and die, so much as that a problem is possible, and easy to prevent, so they might as well require that you take precautions if you go that high.

Upvoted for "They mostly just lose ambition, not their lives."

As many people have noted here, skydivers can start to suffer from hypoxia during the climb to jump altitude.  As per the requirements of §91.211(a)(3) all occupants aboard an aircraft at cabin altitudes above 15,000 ft MSL are required to have supplemental oxygen AVAILABLE to them, though they are not required to use it, except for the required minimum flight crew.  They are required to use supplemental oxygen at cabin altitudes above 14,000 ft MSL (see §91.211(a)(2)).
And many larger skydiving aircraft are so equipped to provide jumpers with supplemental oxygen during the ascent.  Below was a publicity stunt Tom Cruise did for James Corden's late night TV show for the release of one of his Mission Impossible movies.  Here he takes James skydiving and they make the jump above 15,000 ft.  You can see small, clear tubes hanging down along the jumper seat planks in the interior of the Shorts Skyvan they use for the drop.  These are supplemental oxygen tubes which a jumper can place in their mouth and suck on to get oxygen during the ascent.  This is only approved to limited altitudes; anything above FL180 is going to require a full face mask for delivery of oxygen to the jumper.


text:
What is the minimum altitude needed to return to the takeoff airport in a 737 after dual engine failure?
Last week, I had my first lesson in a 737 sim. The instructor, when talking about risks with flying the 737, said the only time it would be "game over", would be if you had a double engine failure on takeoff. He had only ever heard of this happening once.
I asked what the minimum altitude you would need to glide back to the airport and land on a runway would be. He wasn't prepared to give an answer, but when I asked if 15k ft would be enough, he said probably not.
Could anyone answer this for me? Any pilots tried this on a sim and know first-hand?
Cheers!

Wouldn’t that depend on the distance to the nearest suitable airport and the glide ratio? Or is there something special that makes the B737 un-glideable?

Hi @StephenS I'm talking about returning to the takeoff airport

So you’re looking for a rule like 1000ft to make the “impossible turn” like we have for GA planes?

15,000 ft is definitely enough! I cannot answer what the minimum would be, but in this Mentour pilot video they are returning successfully in a full motion sim after dual engine failure at 6200 ft.

@StephenS: "suitable airport", or river, levee, &c.  A lot depends on the definitions of survivable and salvagable.

The instructor was pulling your leg. All depends on what the effective climb gradient is, but assuming you are able to climb anywhere near the gradient in JohnK's answer, 15k is pleeeeeeeenty.

We actually know the answer. The altitude at which Sully lost both engines is the cutoff. +5 more seconds of climb and the return to runway definitely possible. (The simulations say it can be done at his altitude given 2 second reaction time but with the APU off it's not definite that there will be enough hydraulic pressure to come out of the radical turn.)

Most modern airliners have L/Ds of 18 to 22:1 (sorry I couldn't find a direct reference). So from one nautical mile, 6076 ft, in the air you can glide around 18 to 22 nm in still air.
Departure climb gradient at, say 4000 fpm, a typical both engines climb rate, which is about 40kt vertical speed, going 200 kts horizontally, is about 5:1.
So you are climbing at a gradient of 5:1, but can glide at a gradient of 4 times flatter, at 20:1.
So, being able to make it back depends.  In the case of a departure and a immediate climb to enroute altitude, with no level off intervals, you should have no problem returning to the airport after a double flame out at, say, 18000 ft, because you'll only be about 15-20 nm from the airport, and from 3 nm up you can glide 50-70 nm, and you should have ample space to turn around and maneuver.
On the other hand, if ATC (or the SID you are flying) makes you level off for a period at, say 4000 ft, and keeps you there for several minutes, and then you have a double flame out, you will likely be out of range.  If you're only at 4000 ft, your absolute gliding distance is only 13 nm, not allowing for manuevering, and by then you might be 15 miles out. You'll be heading for the river, like those other guys did.

The main problem will be that you're pointing the wrong way, so you also need energy for the turn (which involves pointing your lift slightly sideways, so you're going to lose height there).

Like I said, "space to turn around and maneuver"...

@JohnK Simon didn't say space, they said loss of height.  A 25 degree bank means your lift drops your upward force by 10%.  On top of that, a powered turn is like 24 nmi ( https://aviation.stackexchange.com/questions/8012/what-does-it-take-to-turn-a-747-around-180-degrees ).  And that doesn't get you pointed back at the airport, it gets you pointing parallel to it 15 nmi away.  At 3nmi up and 20 nmi from airport, ignoring height loss from turn (!), it is 23 nmi to turn around and now you are 25 nmi away, so 48 nmi of distance.  I think you aren't gonna make it?  Or did I do a math error?

@Yakk Handy calculator here: http://www.csgnetwork.com/aircraftturninfocalc.html  A 45 deg banked turn (roughly what gives you the lowest altitude loss) done at 230kt has a 1.5 nm diameter, and will take about 40 seconds to turn 180 deg, so I don't know where you got all that.  Say you're descending at 3000 fpm while crankin' and bankin'. You'll come out of a 270 deg turn in about a minute, and will find yourself at about 15000 ft, maybe 22 or 23 miles away, still with 60 miles of potential gliding distance. The problem is getting down, not making it back. Lots of flying around to lose alt.

First, the actual "impossible turn possible" altitude number is going to vary dramatically by factors such as hot, high, heavy, wind, how wind interacts with your departure path, and whatever the damage did to the airplane.
This question  was extensively examined on the very similar Airbus A320, after Chelsey "Sully" Sullenberger had a live-fire exercise on double-engine failure, over New York City, and elected to put it in the Hudson River rather than attempt "the impossible turn" (which would left poor choices in the event of falling short).
Extensive post-mortem study of the accident revealed most pilots in Sully's situation could indeed have accomplished a >180 degree turn back to Laguardia, but only if they had precognition of the extremely unexpected double-engine failure, and started their turn the instant the birds struck.  I won't discuss that further.
The point is: this gives us a pretty good frame of reference: Sully was at 2,818 feet at the time of the bird strike when a return to LaGuardia was possible.
For an A320 in NYC in January... the very opposite of "hot and high".  With the wind Sully had that day.  If Sully's departure had been in the other direction toward Boston, all bets are off.
Further, it would have been a very different picture if an aerodynamic cowling had been knocked off an engine.

That thing is pretty much a barn door. This would have impact on an aircraft's glide range.

I'm no expert, but that engine... that's Not Good™

But nice to see an intact fence between fan bits and pax. Something worked, at least.

@user_1818839 oh, that fence seems to pretty much always work.  The problem is, fan bits won't always throw themselves futilely against that line of defense.  Sometimes they unexpectedly detour via the Ardennes Forest and Belgium.


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What is the best way to deal with a stuck throttle?
Recently I watched a fascinating video of Cathay flight 780.
What interested me about the end of this incident is that the pilots opted to land the aircraft with the engine 1 still at 80%, over speeding with the flap setting (and not even setting full flaps for landing configuration), which was also covered in this question: Why didn’t the Captain of Cathay Pacific flight 780 shut down engine 1 and land with a more reasonable speed?
If it was me, (and I speak as someone who has roughly 1% of the flying time of the captain of this flight and 0% in that type :) ) - I would have got to a high altitude over the airport (maybe FL100) and then cut both engines, so that I could set up a full landing configuration before performing a glide approach.
The benefit of this would be landing at a normal speed. The downside is obvious - you only get one shot.
What is the 'correct' course of action if your throttle gets stuck on full (or close to it) power?

Also note: Flying a huge airliner without engines isn't exactly easy. You lose most electrical power and hydraulic pressure in most systems. There was another incident with an A330 (Air Transat 236, Mentour Video) that lost both engines and ended up with a landing that was very similar to the Cathay one.

After viewing MentourPilot's breakdown of the situation, I imagine the captain did not want to risk the unpredictable engine cutting out partway through a go-around; he figured he had brakes and could throw the functioning engine into reverse thrust upon touchdown, which he did. And he might not have trusted himself with a second Gimli Glider.

Some WWI pilots had to deal with exactly that as normal operation, because the La Rhone rotary engine didn't have a throttle and you controlled the engine with ignition to taxi, descend, and land etc. Whenever the ignition was on, you were balls-to-the-wall.
So just shut the engine down with the mags, keep the mixture on, and glide to where you need to go.  If you need power, switch on the mags on and off as required like you were in a Sopwith Camel. That's exactly what I would do.
You are likely to get nasty backfires when you put ignition on after being off a while, with all the unburnt fuel in the exhaust, depending on how leaky the engine's exhaust valves are, and might even blow out a muffler, but you're saving yourself, so who cares about that.
You could avoid the backfires by using the mixture to do the same thing instead of switching mags off and on, if you can live with the small lag in the input/result. I might use mixture when higher, then use the ignition to manage the engine on approach and cut the mixture for good when the runway is assured.

I presume that technique couldn't be replicated with a turbine engine?

@MikeB No, the ignition is normally off for a turbine engine since you have a continuous flame. The only way to shut it down is to cut the fuel supply. Restarting it also takes significantly longer than a piston engine, plus you typically need a pneumatic source or high enough speed for windmilling.

@jkztd Most turboprops have a free turbine for the propeller gearbox, which would slow down when feathered, but the gas generator would still run at full fuel flow. My guess is you'd overheat and damage the turbine pretty quickly and possibly create an engine fire.

The Vickers FB5 ('Gunbus') was controlled by blipping the mags on the Gnome monosoupape engine.

@Bianfable you would have to leave the power levers above Beta mode (somewhere above flight idle) and feather the prop and the prop load might keep the NG in check (or not). Also, as soon as you retard the power lever to flight idle, you normally enter Beta mode (Flight Beta) where fuel control switches to automatic governing of propeller RPM (Np governing) as the power lever takes over control of blade angle. So possibly a fuel control runaway in flight mode my be resolved by going into Beta, if the Np governing mode still works.

There is no single right answer to this, it depends on the airplane, the number of engines, and the power level of the stuck throttle. Most light singles have no hydraulics to power and your electrical system can run off the battery, if you're in an airliner it's a different story. In a light single with steam gauges half your instruments are usually powered by the engine, if you are in IMC you'd want to keep your engine on as long as possible, in visual conditions you may be more comfortable in a glide approach.
How draggy is your airplane? If I'm in a Cessna 172 with a throttle stuck at cruise power I'd be very comfortable keeping the engine on until I'm on short final because I know I can dirty it up by setting flaps 20 and side-slipping. A Diamond DA-40 is a much cleaner beast as it's design roots are in gliders, energy management becomes much more critical in that situation, I would have to kill the engine much further out in the approach.

The "correct" course of action in any emergency situation is whichever course of action you deem to be most likely to result in the lowest chance of damage to people and property.
If I were flying a light single, which is where I have more or less the only real world experience, and the throttle stuck open I would have almost no option but to line up for a glide approach and cut the engine with by setting the mixture to lean. The reason for this is that I would have no way to land safely with full power - that aircraft cannot produce enough drag to overcome the power of the engine.
In an airliner the situation may be different. A single engine stuck at 80% with the other at or near idle may well be able to excerpt enough drag with flaps/slats/airbrakes to overcome the lift and land without too much chance of killing everyone on board. Ultimately it seems this is the option that has been taken in the example given.

Similar to yourself I have SEP real-world experience and would follow the same approach you detail above, but I'm wondering if there is any possibility of attempting to abuse the mixture control as a poor-man's throttle rather than fully cut the power.  The possible advantage being that you get a little bit of comfort knowing that the engine is not fully cut off, meaning it might be easier to abort and go-around if you think necessary.  I might give it a try in a flight-sim....

@MickWaites bear in mind some aircraft (including the one I fly) don't have a mixture


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What is the difference between RVR and visibility?
What is the difference between Runway Visual Range (RVR) and visibility? Looking at the definitions, both RVR and visibility tells the pilot whether they have enough visibility to land on the runway or not? Are they measured using different instruments or is there something else that I missed out?

The definitions are somewhat different:

Visibility for aeronautical purposes is the greater of:

the greatest safe distance at which a black object of suitable dimensions, situated near the ground, can be seen and recognised when observed against a bright background;

the greatest distance at which lights in the vicinity of 1000 candelas can be seen and identified against an unlit black background.

Runway Visual Range (RVR) is the range over which the pilot of an aircraft on the centre line of a runway can see the runway surface markings or the light delineating the runway or identifying its centre line. (ICAO Annex 3)

(Visibility on SKYbrary)
The RVR is specific to the runway and its actual lighting system (i.e. can be different for different runways of the same airport at the same time), where visibility only refers to the general range an object or light can be seen in the current conditions.

RVR can be measured along the runway, e.g. using a transmissometer:

RVR equipment is typically found at larger airports, and RVR visibilities can be reported for up to four designated runways at an airport. It's calculated using sensors like this:

The sensors are located along the runway, and they're approximately 14 feet higher than the centerline of the runway.
On most runways equipped with RVR, there are three sensors: a touchdown sensor, a mid-point sensor, and a rollout sensor.

(boldmethod.com)
On the other hand, visibility is typically measured by detecting scattered light from a strobe (scatterometer):

The most common method of measuring the extinction coefficient is the forward scatter system. A high intensity xenon strobe transmitter directs a beam of light in the visible spectrum at the air close to the sensor and a photo diode at the adjacent receiver is used to detect the amount of light received. The extent of scattering is increased by the presence of particles of water, dust, sand or smoke. The use of transmitted light within the visible spectrum allows this type of instrument to most accurately simulate human perception of visibility. The range of visibility for which a measurement can be derived is typically between 15/20 m and 20/30 km with an accuracy up to 10 km of +/- 10%. A typical instrument is illustrated below:

(Automated Cloud Base and Visibility Measurement on SKYbrary)
For more details on measurements, see:

How is the visibility at airports measured?

How do visibility and RVR compare?
The ICAO Aerodrome Meteorological Observation and Forecast Study Group published A Comparison of RVR and Visibility Reports:

Previous studies have shown that the reported value of RVR is typically higher than the associated visibility value. [...] These findings are consistent with the fact that the assessment of RVR takes advantage of high intensity runway lights while the visibility assessment is based upon lights of moderate intensity.
[...]
A comparative review of the 5,662 METAR / SPECI from three aerodromes over a five year period that contained both prevailing visibility and RVR information revealed that, in equivalent terms for 2825, or 50 per cent, the RVR report was less than the visibility.
[...]
meaningful predictions of RVR from prevailing visibility reports can only be  accomplished if the assessment of both RVR and visibility is by reference to common sensor(s).

It seems, RVR is typically higher than visibility, but only if the runway lighting system is using high intensity lights.

In practice, RVR gives a better prediction of whether you will be able to see the threshold environment when you get to minimums than visibility, which tends to be more pessimistic.

Indeed, and I've landed many times when "visibility" was below minimums for an approach, but RVR was well above.  This is allowed, partly because RVR sensors are more accurate and better placed, and partly because the RVR scale can more finely describe distance (RVR's minimum unit is 100 feet whereas visibility's minimum unit (1/8SM) is 660 feet).

@nexus_2006 RVR is not more accurate than visibility, it is simply measuring something else. See the answer.

RVR has more precision (increments of 100 feet vs 1/8 mile), which in & of itself isn't the same as greater accuracy, but when the visibility varies between one runway and another (and the point where prevailing visibility is measured), RVR is the more precise AND more accurate indication of your runway's current conditions. It's also, typically, more frequently updated.


text:
What happens with the avionics if in midair there is loss of power or fuel flow interrupted?
What happens if in midair there is loss of power, how will the avionics perform in such a situation (cockpit display, flight plan)?

I'd have answered for smaller aircraft, as you didn't specify size, but since you seem satisfied with the airliner answer ...

You might find this video interesting. It's a reconstruction of what happened when Air Canada 143 (a Boeing 767) ran out of fuel and became the "Gimli Glider"

Does this answer your question? How are fly-by-wire airliners controlled in case of complete electrical failure?

This is somewhat dependent on the aircraft of course, so I will focus on the Boeing 737 NG series as an example of a typical airliner. Details will differ with other aircraft, but the general concept should be the same.
When there is loss of thrust on both engines, both engine driven generators will stop providing power. The battery will take over powering some of the systems immediately. If two batteries are installed, all emergency systems can be supplied for at least 60 minutes:

Two 24 volt nickel–cadmium batteries, main and auxiliary, are located in the
  electronics compartment. The batteries can supply part of the DC system. The
  auxiliary battery operates in parallel with the main battery when the battery is
  powering the standby system. At all other times, the auxiliary battery is isolated
  from the power distribution system. Battery charging is automatically controlled.
  Two fully charged batteries have sufficient capacity to provide standby power for
  a minimum of 60 minutes. Battery voltage range is 22–30 volts.

(Boeing 737 NG FCOMv2 6.20.12 Electrical - System Description)
The batteries power some DC buses directly and the AC standby bus via a static inverter:

Standby Power System - Normal Operation
The standby system provides 115V AC and 24V DC power to essential systems
  in the event of loss of all engine or APU–driven AC power. The standby power
  system consists of:

static inverter
AC standby bus
DC standby bus
battery bus
hot battery bus
switched hot battery bus
main battery
auxiliary battery.

[...]
Static Inverter
The static inverter converts 24 volt DC power from the battery to 115V AC power
  to supply the AC standby bus during the loss of normal electrical power. The
  power supply to the inverter is controlled by the standby power switch and the
  battery switch on the overhead panel.

(Boeing 737 NG FCOMv2 6.20.16 Electrical - System Description)
The FCOM contains a long list with all systems that are powered in this configuration. In general, only the captain's side instruments will be available. I will not quote the full list here, since you only asked about "cockpit display, flight plan". The relevant items are:

All Generators Inoperative
The following list identifies the significant equipment that operates when the main
  battery and the auxiliary battery are the only source of electrical power.

emergency instrument flood lights
Captain's pitot probe heat
VHF No. 1
upper display unit
Captain's outboard and inboard display units (EFIS/MAP or PFD/ND)
clocks
left EFIS control panel
Standby instruments
FMC
left CDU
heading/track indications
VHF NAV No. 1
ILS No. 1
left IRS
left GPS

(Boeing 737 NG FCOMv2 6.20.19 Electrical - System Description)
This allows the Captain to continue flying the aircraft via manual reversion (no hydraulic pressure will be available, no autopilot is available). An emergency descent will have to be initiated without thrust. The checklist in the QRH will aim at restarting one or both engines. If that fails, the APU can be started in flight below 25,000 ft to provide more electrical power (all AC buses powered, all displays available again, electric hydraulic pumps available again).

As an additional piece of information on the backup batteries - batteries were initially a big problem for the 787 Dreamliner as they were Lithium-Ion and had a few early situations with thermal-runaway of these advanced-technology batteries.  (https://en.wikipedia.org/wiki/Boeing_787_Dreamliner_battery_problems).  I'm not aware that the danger was completely resolved - according to my reading, they just beefed up the protection in case of a battery failure.

Isn't it possible to get some engine RPMs from windmilling, and will that get generators working at all?

@Harper-ReinstateMonica Windmilling RPM is typically around 11% N2, which is not enough for the generators to work. I cannot find a source for the RPM at which the generators start working, but usually they are not used below 56% N2, which is where the starter cuts out.

This post is missing the crucial lift over drag ratio, and how it corresponds to maximum distance an aircraft an travel.

@paul23 Not sure how that relates to the avionics OP asked about. The assumption is always that you can eventually restart at least one engine. If I understood the question correctly, OP asked about how the avionics is powered during that time.

@Bianfable during my education for aerospace engineering designing for landing/gliding to airport was quite standard when you design an aircraft. "Your engines have both failed, and rudder has a maximum deflection due to damage, what is the maximum crosswind the aircraft  you designed can land at" is one of the standard exam questions in the first years.

Although the accepted answer is basically correct, as an additional backup source of power (for hydraulics or electric, depending on the aircraft), there can be a Ram air turbine (RAT - https://en.wikipedia.org/wiki/Ram_air_turbine) that can be deployed in a failure of both primary and secondary energy sources.  It uses the airflow from the airplane's speed in the air to generate energy for hydraulics or electrical needs.

Nice addition! The 737 does not have a RAT, so I didn't mention it, but most modern airliners do (basically everything fly-by-wire will have one).

The first aircraft I saw this fitted on was the F-4 Phantom II which did not have battery backup for primary systems at least for the K and M variants (the only battery in the aircraft was there to light up the rear cockpit eject light).

Actual details vary between aircrafts. But in general, the battery can supply essential systems for some time and then APU/RAT/... should kick in.


text:
Is it possible for an airplane to enter another airplane?
I was making some cool fictional airplane designs and realized that most of my larger airplanes have a bay inside that stores a plane or two. But then I thought, is it even possible to make a plane exit and enter a plane mid-flight? Such a plane to enter another plane would need to have folding wings to take up the least space when stored, and also because the plane it is entering would have to be significantly larger to have a fuselage wide enough to fit a smaller plane's entire wingspan. Note that I am questioning if it is possible to do something NOT in an emergency, but as a tactic or for other reasons. That being said, the question differs from an emergency situation due to the fact that the question is implying that the operation be done over and over, and not in a risky, dangerous, last-resort fashion.
This mainly branches two questions:
Is it possible for a plane to exit a plane mid-flight, and if it is possible for a plane to enter a plane mid-flight. Clearly the plane must enter and exit through the belly of the larger plane, because something like a large hydraulic arm would be necessary to "grab" the small airplane and allow it to fold its wings up before hoisting it into the plane. This came to mind after considering that the plane must folds its wings up before entering the plane, and it also must take at least 5 seconds to fully fold the wings. So, folding the wings mid-flight independently isn't a good idea: The plane will plummet without wings to lift it. And to exit a plane, I'm assuming the arm will have to hold the plane out of the bottom and allow it to fold the wings out before dropping it. But here's what challenges that idea: When the arm is holding the plane and plane's wings are extended, will the little plane's wings effect the flight of the Larger plane? (Since the arm is grabbing it, it is the equivalent of the large plane having two tiny wings below it) and also, what are the effects of dropping a plane mid-flight? Any help on this would be very, very appreciated.
Oh, and a side-question from the above: Would a plane large enough to fit another plane be large enough to reasonably carry the weight of that smaller plane? And if so, how will the larger plane fair considering the fact that such a hydraulic arm necessary for mid-flight entry and exit of the large plane must be considerably heavy?

Not quite. That question has answers that would differ significantly from this question. See, that question literally specifies it as a last-resort, an emergency. Another critical difference is that I am also asking if it would be possible to exit the plane as well. It might also be helpful to consider that both these aircraft would likely be designed to do this maneuver repeatedly.

Some post WW2 bomber designs had parasite fighters; the fighters would be deployed by a trapeze, do their thing, and come back to hook on. It didn't really work.  https://en.wikipedia.org/wiki/Parasite_aircraft  (I just looked at the wiki and the concept  pre-dates WW2!) Looks like the first picture in the wiki article has your "hydraulic arm". It kinda sounds to me like you've reinvented the parasite fighter.

@OrganicMarble yeah, the Soviets and Americans both experimented with the concept prior to WW2, ultimately unsuccessfully.

Ginger,look at https://sacmuseum.org/project-ficon-and-the-b-36/ for a real world example that kinda worked but never entered production

Could be a fun remake of "The Italian Job"...

I bet there's something like this with drones.

@user_1818839 - as distinct from the garbage remake someone made a few years ago !

Clearly the plane must enter and exit through the belly of the larger plane

Why so? If you drop this condition, things become much easier, because you could attach the smaller planes externally and avoid folding the wings. And this was indeed tried and even used operationally.
The concept is generally termed as Airborne aircraft carrier. The first (?) and most obvious application of the idea was with airships, where the airplanes could indeed be stored internally.

Wiki
Russians extensively tried the idea in the 1930s with up to 5 fighters attached on or under the wing of a large bomber.

Wiki
Admittedly, this configuration was mostly indended for deployment of the fighters, which would then land normally. This way it was used several times successfully for bombing raids early in the WW2. Recovery of fighters was tried, and some configurations even assembled in the air rather than on the ground. But it was quite difficult and required considerable skills even without any folding wings.
During the tests, on some occasions, there were problems (including at least one fatal incident) you are worried about, namely negative influence of the fighters on the carrier. Mostly they were related to failures, when the attachment broke down or didn't work as expected (sometimes due to incorrect control of the fighter during the process).
However, even pretty much the same idea as yours was explored in the 1970s, with the B747 or C-5 as the carrier and some "microfighters" as the payload. Apparently, the concept was deemed "feasible" but never went to an advanced stage of development. Yet today, the idea may be of some interest again with UAV aircraft.

Wiki

The 747 AAC has a touch of the Marvel Cinematic Universe to it!

The concept has been used with UAVs mounted under the wings of C-130s during the Vietnam war. These were however not recovered in the air but flown to friendly occupied territory and landed (usually crash landed, as they had no landing gear).

The aviation channels Mustard and Dark Skies have numerous micro documentaries about experimental parasite aircraft e.g. https://www.youtube.com/watch?v=WE4Uxo6WLio
https://en.wikipedia.org/wiki/Parasite_aircraft
Also, if you like the 747 AAC you should look up Japan's WWII underwater aircraft carrier!

the people designing the 747 AAC must have watched too much Battle Star Galactica!

For another nice example see Mercury and Maia, the Short-Mayo Composite. Use case was a range extender to allow a mail plane to cross the Atlantic. (Carrier launches mail plane inflight and returns to base, presumably around 1/3 of the way across. No inflight landing possible. https://en.wikipedia.org/wiki/Short_Mayo_Composite

"The 747 AAC has a touch of the Marvel Cinematic Universe to it!" — Could you explain the "Marvel Cinematic Universe" part?

It was actually done post-WW2 as part of the USAF FICON and Tip Tow projects which thought to extend the service life of the B-36 by giving it some escort fighters and photo reconnaiscance aircraft for damage assessment.
It ultimately wasn't put into production as it was rather dangerous and seriously degraded the performance of the carrier aircraft, meaning they'd be incapable of keeping up with the already slow bomber formations.
http://www.air-and-space.com/ficon.htm shows a lot of details of the actual system as built into the B-36 carrier aircraft, and photos of the parasite fighters involved.
"In parallel to both FICON and TIP TOW, project TOM TOM was being developed using a JRB-36F and two RF-84F. Using a newly designed articulated arm and clamp system, the RF-84F was intended to attach to the JRB-36F and be towed to the target site. The JRB-36F retained most of its cameras to record flights and the RF-84F would retain four of its .50 guns for defensive action for itself and the “Mothership.” Turbulence and vortices continued to be a problem for the RF-84F pilots, with one aircraft actually being torn from the JRB-36F in-flight. The program, much like TIP TOW was cancelled." from https://sacmuseum.org/project-ficon-and-the-b-36/ gives some details of the problems encountered.
In the end it was decided to retire the B-36 and build more B-52s instead.
For non-military missions, there were also the X-1 experimental rocket planes carried aloft by B-29 motherships and X-15s by B-52s, but those were incapable of being retrieved in flight.
enter link description here lists some other projects with similar concepts that actually flew.

Consider a C-5 Galaxy and a T-38.
Big and little.
The C-5 cargo compartment is 19 feet (5.76m), The T-38 wingspan is 25 feet (7.7m). So not going to fit.
Weight capacity of the C-5 is 291,000 lbs, and the little T-38 onl weighs 12,000 lbs or so. So THAT isn't the problem.
Now...if you were to park the little jet on a platform extended from the back, and then fold the wings before pulling it inside, sure.
The question is, outside of fiction...WHY would you do this?

there are multiple benefits that it has. If you're flying a large plane and you're fighting enemies, and the enemies are tiny, zipping around all over the place, not an ideal match for a large plane. Sure, the large plane is stronger, but it just sits there, being an enormous target. Then you could switch planes to a smaller plane suited for dogfighting.

How would you switch planes? It'd have to be inside of your plane. Because you can't just swap with an airplane flying right next to you. That's risk and dangerous. And if this is a long-distance mission, the little planes couldn't come as escorts because they wouldn't be able to travel such a distance on their own. It can also serve to surprise the enemy. If the enemy doesn't know you have an escort because it is IN you, then they couldn't possibly plan ahead and attack you with the planes suited for fighting escorts, now could they?

Outside of fiction (that would not be believed), air combat does not work like that.

@Ginger how would you get into the parasite if it's (partially) external? With some difficulty and acrobatics. As was shown by the test pilots of the USAF FICON project, and the pilots of some of the later X-1 derivatives who'd climb through the bomb bay of the carrier aicraft, lower themselves into the parasite aircraft's cockpit, and in reverse during longer flights. And no safety harness, if they slipped it was a long way to the ground below with no parachute.

@WPNSGuy yet the USAF seriously investigated it for their B-36 fleet to keep them relevant in the face of Soviet jet fighters capable of intercepting them. It was deemed too dangerous and never put into production, but several (I know of at least 3 just for the B-36) attempts were made.

@WPNSGuy, don't consider the T-38, consider the Arado E.381, or the McDonnell XF-85, or the Martin Marietta X-24A.  There are some seriously short-winged aircraft out there.

@Mark - The Goblin is still a bit too wide. The X-24 would fit (barely), but its not really big enough to carry any weapons. So if we were to design a carrier aircraft, AND the parasite aircraft...sure, this could be done. But given a 20 year lead time to deployment (2040+), we'd be looking at unmanned aircraft. Make those any size you want. Given current aircraft? No.

Sure, it's possible. There have been several experiments with so-called "parasite aircraft" over the years, some of which even managed some success before inevitably being canned for reasons of practicality and cost.
Docking them together is going to be the hard part. A hydraulic claw won't work; aside from the sheer weight of such a device, planes bounce around way too much to have a realistic chance of grabbing one mid-flight.
I'm picturing something similar to current midair refueling equipment. The big plane would trail a cable that engaged with a (deployable?) hook on top of the small plane. As the little one shut down its engines and folded its wings, it would naturally fall beneath the big one, and could then be winched in. Still not remotely practical, but believable enough for a fictional world.

Will the little plane's wings effect the flight of the Larger plane?

Yes, but only slightly. Any reasonably competent pilot could compensate easily.

What are the effects of dropping a plane mid-flight?

The big plane will suddenly be a lot lighter. The pilot would need to push forward to avoid climbing (although, come to think of it, they probably want to let it climb, at least a little, to reduce the chance of a midair collision). It would, in fact, be very similar to dropping a bomb, so you could look up info on big bombers for inspiration.

Would a plane large enough to fit another plane be large enough to reasonably carry the weight of that smaller plane?

Shouldn't be a problem. After all, big cargo planes carry multiple fighters to deployments all the time.

How will the larger plane fair considering the fact that such a hydraulic arm necessary for mid-flight entry and exit of the large plane must be considerably heavy?

As mentioned above, a hydraulic arm would be the least likely part of the little scheme, which is why I recommend cables and winches instead.

Landing a plane inside another plane would entail the airspeed falling to zero, which would be possible during parabolic (zero-g) flight, although the control surfaces would become inoperative, so some kind of thrusters might be required.  Failing that, if the larger aircraft were open at both ends it would be rather easier.  I’ve seen paragliders deploy from helicopters and microlights; in principle many types of aircraft could be deployed in flight provided that they can stay within their safe envelope while adjusting to the airflow.

no parabolic path needed, you "just" need to perfectly match your airspeed with that of the carrier. Some sort of trapping mechanism would almost certainly be deployed to kill the last bit of relative airspeed.

Putting the carrier aircraft into freefall really doesn't make anything less complicated.  It not only makes it difficult for the landing aircraft to land, it imposes hard time constraints on the approach.

Having no relative airspeed is ok until you suddenly have no drag and must therefore reduce thrust to zero, and have no lift.  That’s not ‘just’ anything.  While the mathematics may be no more complicated, the reality is that a straight and level approach has far less margin for error than a parabolic path.


text:
Could drag in a turn be twice the drag of that encountered in straight and level flight?
I Just wanted to see if my understanding of drag is correct, as I am unsure.
I understand that during straight and level flight, the drag can be calculated as:

With the first term being the parasitic drag component and the second term being the induced drag component.
However, in a coordinated turn with the same angle of attack at the same airspeed, I have been told that the total drag can be expressed as:

And because the angle of attack is the same, the lift to drag ratio is the same. In this scenario the load factor is 2.
However, since in straight and level flight L = W. So does it follow that the total drag in this turn is double the total drag in the straight and level flight?
i.e.

I don't know why but I feel that this is wrong.
If this is correct, however, the question says that the parasitic drag of the turn is equal to the parasitic drag of the straight and level flight; so would it correct to follow that the induced drag is equal to the following equation?

You should clarify whether you mean the same airspeed in both cases.  If you do, there is an error in your application of the "same angle of attack" idea.

Hi, I've edited the question to reflect that the airspeed is the same in both cases. I thought that the same angle of attack maintains the same lift to drag ratio in turning and straight level flight

In "a turn" with a 10 degree bank angle, the drag is probably well less than that of level flight. If you're in an F-16 with a 9-G turn at a bank angle close to 90 degrees, then your drag is well beyond double the level flight case.  Somewhere between those two extremes, you'll find "a turn" (quoting your question's title) that gives you 2x the drag of level flight.  Is your question if that always occurs at 60 degrees bank & 2 G's?

yes in this question its 60 degrees bank with load factor 2 G's

@Ralph J in "a turn" with a 10 degree bank angle, the drag is probably well less than that of level flight???

What would really neat is to hold the aircraft level and turn using rudder and fuselage as a symmetrical airfoil, generating sideways lift.  But, a little more drag there too.  (My original thought was to turn just using the wheel spats).

Sorry, 10 degree banked turn, drag well less than double the S&L case. The 9G turn has drag well above double the S&L case.

While airborn lift always equals weight, doesn't matter what sort of contortions or altitude changes are happening. Weight is a shorthand for net force, it is not mass, don't confuse mass and force.

Could drag in a turn be twice the drag in straight and level flight?

Yes, but this will be a rater steep turn.

…  in a coordinated turn with the same angle of attack at the same airspeed …

No, this is impossible. Either angle of attack or speed have to be higher so the higher lift required for turning can be generated.

… angle of attack is the same, the lift to drag ratio is the same. In this scenario the load factor is 2

And in order to create twice the lift of straight flight, the speed has to be higher by a factor of √2. In this case, L/D is indeed the same, but speed is higher so twice the dynamic pressure is available for creating the required lift. Since both lift and drag are proportional to dynamic pressure, creating twice the lift at the same L/D means that also drag will double.
In order to turn with a load factor of 2g, the bank angle has to be 60°. That is rather steep.

For a set air density, a given airspeed and angle of attack will always produce the same lift and drag.
lift = weight in all cases and Weight=Mass*Acceleration(gravitational and mechanical combined. Vector sum.) If lift=weight and weight=M*A then lift=M*A and if M=1 and A=1 then L=1, if A=2 then L=2. If we know L=1 based on air density speed and AoA then W=1 and if M still =1 then then A must =1. The other option is that A=2 and so M=1/2.
Although load factor is tossed around quite often in aviation, even by many very competent instructors, I don't feel it is used appropriately in most cases. "Design load factor" is a circuitous method of specifying a target stress condition of an airframe for baseline engineering purposes.
G units of acceleration is really more accurate terminology for lift vector calculations in my opinion. one G unit = about 9.8 in kg/M/S units or 32.2 in Slug/foot/S units
Alternately you could just use weight regardless of its source for your calculations. That is the lift needed for 2*mass at 1G is the same as 1*mass at 2G, and aerodynamically equivalent. With the possible exception of an extremely small radius turn where you would get effects from the tail and nose experiencing significantly different angles of attack, but this would be a radius of only a few fuselage lengths.


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How many years can a Boeing 747 be in service?
Over the years, many 747s have been retired from passenger service, but some are still flying in other roles, like cargo transport. So, how long can a Boeing 747 remain in active service before it is retired or replaced?

Keep in mind that in addition to the excellent answer regarding strict regulatory requirements for inspection and/or reinforcement, a new plane of comparable capacity will have a much lower cost to fly due to using two newer engines instead of four older engines.

The 747 is has a design life of 35000 flight cycles (takeoff and landings) as it came from the factory (this could vary with model). Around 150,000 hours give or take.
Once you get to that level, you will be faced with a couple of options; reinforcing the structure to maintain the existing inspection intervals, or carrying on with much more frequent inspection intervals.
A lot of the structural inspections will require Non Destructive Test equipment like Eddy Current (which can find hidden cracks), and the expenses involved go through the roof if the intervals have to be reduced to much lower values to maintain fatigue safety margins.
The alternative may be to install reinforcements (sold by Boeing or a 3rd party as service bulletin or STC kits) to beef up the structure and restore it back to or close to the original inspection intervals.
One option involves recurring financial pain supporting frequent structural inspections, the other involves one big bout of financial pain to buy, say another 20,000 cycles of operation under the old inspection program by purchasing a reinforcement kit.
At some point, the business case no longer works for either option and it's just not profitable to operate it, and the airplane is put out to pasture, usually stripped clean of parts for resale or re-use, and the airframe sold for scrap.

It will depend on the length of the routes but 35000 cycles/ 150000 hours comes to something like 40 to 50 years for a typical use case?

A 747 average leg will be somewhere between 5-8 hours. Max endurance is well over 10. Flight cycle sensitive areas are landing gear structure, fuselage swelling with pressurization, that sort of thing.  If one does two legs/day. Maybe 500-1000 cycles per year. You could say 30-40 years is typical before the airframe is becoming too expensive to keep up. That's if corrosion doesn't do it in first. The inside of a pressurized fuselage gets very damp because the insulation gets very wet from condensate forming on inside of the outer skin. This ends up near the bottom and can rot the structure.

and even before corrosion or flight cycles make the aircraft uneconomical (or worse, unsafe) to operate there will be a time when operating cost due to consumables (fuel, oil, etc. etc.) starts to outstrip that of newer aircraft to the point that scrapping the 747 in favour of say a 777 freighter starts making sense (of course that also depends heavily on the operator, someone like say IranAir won't have easy access to replacement aircraft).

Yes.  Fuel price fluctuations are a huge factor.  Operators were retiring CRJ200s before their airframe lives were used up, because they were becoming too expensive to operate, but when fuel prices imploded following 2008, it gave them a new lease on life. Skywest was snapping up all the used 200s they could get their hands on.

In 2011, FAA and EASA adopted coordinated rules against widespread fatigue damage. At some point, large portions of the structure have reached their fatigue lifetime. Multiple cracks may form and start to merge and the whole airplane will blow apart, like Aloha Flight 243.
Thus, pressurized transport aircraft have had hard limits on flight cycles and hours since 2011. Any aircraft exceeding those limits loses its airworthiness certificate. Aircraft under the Soviet/CIS system have had such limits for many decades, this was finally adopted this in the West.
For the 747, there is a default limit of 20,000 flight cycles. Boeing has extended this limit to 35,000 flight cycles and 165,000 flight hours (-400, 135k FH prior models) with inspection and modification work.
This is a hard limit, so aircraft reaching that limit cannot be operated in FAA, EASA, and countries following their rules. Manufacturers will also end support at that time. The aircraft must either be scrapped, or it may find itself in countries with far lower aviation safety standards.

Interesting. That seems to contradict the other answer to a degree ("hard limit" is different than "install a reinforcement kit"). In any case I'd be curious to know whether this just affects commercial operation. After all, enthusiasts fly vintage airplanes essentially "forever". At some point it may be a "ship of Theseus", sure, but with enough love and care, shouldn't it be possible to keep even a large vessel airworthy (for non-commercial operation)?

@Peter-ReinstateMonica: Due to the square cube law, replacing a strut on a vintage aircraft with an unreasonably strong one is snot likely to put the plane overweight.

@Peter-ReinstateMonica The LOV requirements on recent planes are written in Part 25 and the type certificate, so at expiration of LOV, the aircraft cannot be maintained to the TC. The avenue there would likely be to come up with an inspection regime and operate the aircraft under an experimental TC, which prohibits commercial use. Vintage aircraft have done this when major repairs are needed. For older planes certified before this rule, it's Part 121 that bans it, so non-air carrier operation may be possible, though it's also been deemed unsafe, so again, going to an experimental TC is likely.


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What is wing trammelling?
What is wing trammelling and how does it affect wing stability and performance?

It's just making sure the rectangle formed by the spars, compression tubes and cross bracing is square.  It's done using a trammel bar, which is just a stick (which the guy in the picture is holding) with a metal point at each end that is used to make sure diagonal reference points along the spar beams are the same distance apart when set across one leg of the X formed by the cross bracing, then the other leg.  The effect on stability and performance is indirect insofar as you don't want a rectangular wing to be trapezoidal (swept - unless the sweep is part of the design like the upper wings of some biplanes, in which case you trammel to that requirement), as well as the fact that parts won't fit if things aren't square.

Pls pardon my ignorance,........so the purpose of wing trammeling is to make sure that the internal braces form squares in a rectangular wing,correct? @Jonh K

Yes, in a perfect 90deg rectangle the distance between to opposite corners is the same. A trammel bar usually has a point that can be slide to a specific distance, such as the center of two wing spars at given rib stations. The opposite rib positions should be the same distance - if not then you have a parallelogram, not a rectangle. see https://en.wikipedia.org/wiki/Parallelogram


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How much g-force is experienced in a 45° turn?
A 60° turn in an airplane doubles the g-force; how much g-force is experienced in a 15°, 30° and 45° turn?
Is it 0.5, 1 and 1.5g respectively?

It depends on your climb or descent rate during the bank.

It can't be 0.5 g, when it's already 1 g in straight flight.

It also depends whether you're upright or inverted - and also the altitude (what's the G-force in a 45º bank in an SR-71 at 85,000ft?)
Certainly assumptions can be made about the question or asserted in the answer, such as upright, level, coordinated flight, in an airplane, below 40,000ft.

Upright/inverted is, generally, contained in the "bank angle."  The altitude doesn't matter to the load factor until you're not actually flying anymore.  Coordinated/level does matter.

I read it the arc turned, so keeping all equal (bank, radious, velocity, attitude, climbing, etc) the experienced g can be the same (just for more time)

@hemp Considering that gravity at 400 km above the surface of Earth (for example, as experienced by the International Space Station) is about 90% of that on the surface, and gravity falls as distance squared, I think it's safe to assume that the gravity at 25 km is, for our purposes, equal to that on the ground. It certainly is close enough to its value on the ground that the difference will be of mostly academic interest. Orbit isn't high up; orbit is fast.

@MichaelKjörling The question is mostly of academic interest, and the intent of my comment was to point out how pedantic many of the other comments were (apparently missed the joke). That said, I did check and gravity at 25km is approx 9.7239 m/s2.

@ErinAnne if you are inverted in level flight, your bank angle is zero, so I'm not sure how you think that's contained. Altitude does matter, it affects the force of gravity from the earth. A 45º banked turn on the moon would be a lot different, ask an astronaut. It is a rounding error in practice, for sure, but the question didn't indicate the level of precision required to answer it. Of course there may be a formula for "relative g-force" which would be unaffected by actual gravity.

@hemp in this context, and many others, it would make more sense for inverted-level-flight to be 180 degrees roll/bank, not zero.  In a terrestrial context, load factor is already relative to the lift needed to loft the airplane, so constant gravity need not be assumed.  And yes, a 45 degree banked turn on the moon would be different, since you wouldn't be generating lift using a wing and so none of this applies at all.  Your humor is not only not coming through but it distracts from a worthwhile note that assumptions matter.

"it would make more sense for inverted-level-flight to be 180 degrees roll/bank" .. I'm not sure it makes sense in any context. The G-Force experienced by the pilot in level, inverted flight is 1G. It happens to be a negative G versus a positive 1G, but it's still 1G. The load factor experienced on the airplane depends on the airframe. In most non-aerobatic aircraft, it is higher when inverted than while upright due to bias in the elevator, but the question was about G force, not load factor.

The formula for the load factor $n_z$ in a turn is
$$n_z = \frac{1}{cos\phi}$$
where $\phi$ is the bank angle. This makes the load factor in a 45° turn 1.414 or $\sqrt{2}$.

Since Michael correctly mentions the simplifications in his comment, let me expand the formula above. In case of a climbing turn, the load factor depends on the coordinate system used. If it is measured in airframe-fixed coordinates, the load factor will be split into a z-component ($n_z$, pointing down) and an x-component ($n_x$, pointing fore- or backward, depending on the aircraft's pitch attitude $\Theta$; nose-up is positive):
$$n_x = -\frac{sin\Theta}{cos\phi}$$
$$n_z = \frac{cos\Theta}{cos\phi}$$
For an uncoordinated turn the value needs to be modified further with the relationship between bank angle change for a given sideslip angle, which is aircraft-specific.

This is true only for level, coordinated flight.

@MichaelKjörling is it for level flight, or is it unaccelerated flight?  I think the same applies to e.g. a steady-state climb, we just don't think of doing that sort of thing as often.  (I typically also think coordinated/level for these but I just got to wondering).

The G forces (also known as the load factor) experienced by an aircraft and its occupants do not scale linearly with the bank angle of the aircraft.
Assuming constant vertical speed, coordinated flight (unchanging rate of climb or descent, ball centered), the load factor at a bank of 0° is 1.00. As the bank angle increases, this grows – slowly at first, then faster. At 15° the load factor is about 1.03; at 30°, about 1.15; at 45°, about 1.41. Increasing bank angle further, at 60° it's 2.00; at 75°, about 3.86.
In general, assuming level and coordinated flight, the load factor at a given bank angle $\theta$ is calculated as $$ n = \frac{1}{\cos{\theta}} $$
For example, for a bank angle $\theta = 45°$, we have $$ n = \frac{1}{\cos{45°}} \approx 1.41$$ and sure enough, the load factor for a level, 45° bank is just that.

Using three significant digits it is 2.00, using four it is 2.000 and so on...

@DaG Sure, but 60° has at most two significant digits.

It was just to say that, since that is an exact value (and a simple one: 2), it sounds a bit peculiar to mention how many significant digits are we considering.


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How do I calculate the lift force on an aileron?
How do I calculate the force of lift acting on an aileron with a deflection of 8° and a flight speed of 40km/h?

How big of an aileron? How big of a plane?  40 khm is just under 25 mph. On my Cessna Cardinal, which has ailerons on the outer third of the wing, that little aileron movement at that speed while just rolling
(takes 60-65 mph to get airborne) would have very little affect.

Cessna Cardinal planform, I don't know how to put this in a comment

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Why are wings load tested upside down?
Why are wings load tested by putting weights on the underside of the  wing (placed upside down)? A wing is being pulled into the air from its upper surface in flight, so shouldn't it be load tested in the other direction?

No, a wing is not being pulled up by the low pressure on the top side. NASA. Wikipedia.

It can also be tested by pulling it upwards.

So is it fine to hook ropes to individual ribs and hang the plane in the air with the gross weight in the plane inorder to load test?@foot

@MichaelK There's no low pressure on the top side? Really? https://aviation.stackexchange.com/questions/19241/why-does-condensation-form-on-the-wing-especially-during-take-off-and-landing

@PhilFrost Did not say that.

So there is low pressure but it's not pulling the wing up? How's that work? What's lifting the plane, if not something pulling up on the wings?

@PhilFrost read the NASA link MichaelK posted above, keep clicking next and the final page explains it. Alternatively try this question.

I get that the equal transit theory is incorrect, I just don't see why @MichaelK is jumping to that, given what's in the question. The question doesn't seem to say much of anything about how lift is generated, just that "a wing is being pulled into the air from its upper surface in flight". Is not the pressure on the top surface lower than the pressure on the lower surface of the wing? Does this not mean there must be a net upwards aerodynamic force on the wing, "pulling it up"? How does this invoke the equal transit theory?

@MichaelK Your link directly contradicts your claim!  Read the last bullet!

I think what @MichaelK is getting at is that if there's lower pressure above the wing, the wing is not being pulled from the top, but pushed from the bottom.

MichaelK is incorrect. The wing has lowered pressure on the top which provides the majority of lift and an increased pressure on the bottom which provides the minority of lift (both lifts in comparison to the pressure before the wings). It's just that the equal transit time idea is wrong.

I think it's most accurate to say that the pressure differential between top and bottom results in a net lift force on the wing.  It also results in a mass of air being thrown downwards to balance, which ultimately results in wingtip vortices.

@ThorstenS. MichaelK is technically correct, but hasn't clarified the misunderstanding. While there is definitely lowered air pressure on top of a wing, air doesn't "pull upward" on the wing. Fluids can't ever pull anything, they can only push. Lift is generated by the side with high pressure pushing upward. This lifting push is counteracted by pressure on the upper surface pushing downward. Thus, as Chromatix said, the bigger the difference in pressure, the greater the lift. (This all assumes both surfaces have the same area, as force is pressure times area.)

Or in other words, the secret to getting an aircraft to fly is to stop the air on the upper surface of the wing from holding it down.

A wing can be tested in any orientation as long as the load is applied correctly. The classic wing test photo is the 787 in a fixture showing its extremely flexible wings.

I thought it might be fun to add an ultimate load test, so here is the 777 tested to failure. Sorry for the early 1990s video quality.
Boeing 777 wing fail at 154% design load

@PilotHead is correct, but to elaborate a bit on why traditionally weight was put on the bottom is largely because its just easier. If you are Boeing you can afford to build a rig large enough to hold a plane down while you pull the wing up. If you are building a home build in your garage its far simpler to put some sand bags on it and let gravity do the rest.

Interestingly, this sand bag example shows how not to do the test as the wing is not inverted and I doubt the intent was to test negative Gs. The tail was tested the same way which is ok since it is intended to provide download.

@Pilothead I did notice that as well but its a good picture outlining the process. Lower down on the page they are testing it upside down

@Pilothead, note that you do have to test some negative Gs, as even the category with least requirement is to be tested from -1 to +2.5 G.

@JanHudec The photoset in question had twenty photos showing only one wing orientation, and it is almost impossible that they would only test for negative G. A different group did the test correctly.

Think about the forces on an airplane (free body diagram). Where does most of the lift come from (wings)? Where is most of the weight(fuselage)? That means that the wings are pulling the fuselage up against gravity (or the classical way of thinking about it, the fuselage is pulling the wings down)

As far as I know, all aircraft have structural limits for both positive G and negative G (typically a lesser value) that need to be proven. A small aircraft may be designed for loads of positive 5G (wings pushed up relative to fuselage), but only negative 3G (wings pushed down). The well-known image of fighter aircraft quickly descending by steep banking first is an illustration of this difference in structural performance (as well as visibilty concerns). The aircraft can withstand quite high loads when suddenly pulling out of a dive, but usually much lower loads when suddenly departing level flight into a dive.

Pressure always pushes.
Low pressure doesn't pull.  Low pressure just pushes less than high pressure.
When you have a stretchy fluid-filled pressurized sack (like a human), its internal pressure causes it to really want exterior pressure on it.  When you don't, you can have fluid leaking or the stack tearing itself apart.
Similar problems can happen with metal holding back pressure.  But even there, the force comes from the air inside the metal object pushing more than the air in the low pressure zone.
Vacuums don't suck things.  It is the air pressure surrounding is that is violently shoving things into the lower pressure zone they create.
It is true that you don't get the same forces from upside down weights that you do from a wing in flight; but there is always going to be a mismatch between your model and your test.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

@garry the OP asked why they are being tested by pushing when they are pulled in flight; I explained that pressure only pushes, and that they are wrong in their belief that pressure pulls.  Does anyone really need a sentence that connects "pressure actually pushes" to "test by pushing is appropirate"?

I'm not saying you're wrong, just that you're not really answering the question.  Your statement would be a good comment adding to Dave's answer. And as the Boeing example shows, it can be tested by pulling up on the wings. Ultimately, it doesn't matter whether you push or pull or the wing is right side up or not, as long as the test applies the proper load.

It seems to me that if the entire premise of a question is wrong, correcting the premise is a valuable answer. It's OK to test wings by attaching cables to them and pulling despite the fact that air lifts the aircraft by pushing on the wing, not because air pulls on the wing. I would only suggest that the answer should focus on top surface vs. bottom surface pressures on a wing, not inside vs. outside pressures on skin or the walls of containers.

@DavidK I added that to explain why it feels like sucking and not pushing when you put your hand on a vacuum cleaner hole.  The "sucking" is really just the fluid in your hand pushing out of your body and your skin being stressed by the lack of support.